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aleksandr82 [10.1K]
2 years ago
7

The arm of a crane at a construction site is 17.0 m long, and it makes an angle of 11.6 ◦ with the horizontal. Assume that the m

aximum load the crane can handle is limited by the amount of torque the load produces around the base of the arm. What maximum torque can the crane withstand if the maximum load the crane can handle is 643 N? Answer in units of N · m.
Physics
1 answer:
riadik2000 [5.3K]2 years ago
8 0

Answer:

τ_max ≈ 10,708 N.m

Explanation:

To find max torque, we need the component of force that is perpendicular to the direction of vector r.

Now, Torque is the perpendicular component of the force times the distance.

Thus;

τ_max = F x r

τ_max = (643 cos 11.6°) (17 )

τ_max ≈ 10,708 N.m

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Answer:

Option B

Explanation:

The phase difference is found by subtracting the 2.3m for the receiver from the other speaker which is 2.9m hence

Phase difference= 2.9-2.3= 0.6

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A teacher uses the model that little invisible gremlins speed up or slow down objects and the direction they push gives the dire
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2 years ago
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A p-type Si sample is used in the Haynes-Shockley experiment. The length of the sample is 2 cm, and two probes are separated by
Airida [17]

Answer:

Mobility of the minority carriers, \mu_{n} =1184.21 cm^{2} /V-sec

Diffusion coefficient for minority carriers,D_{n} = 29.20 cm^2 /s

Verified from Einstein relation as  \frac{D_{n} }{\mu_{n} }  = 25 mV

Explanation:

Length of sample, l_{s} = 2 cm

Separation between the two probes, L = 1.8 cm

Drift time, t_{d} = 0.608 ms

Applied voltage, V = 5 V

Mobility of the minority carriers ( electrons), \mu_{n} = \frac{V_{d} }{E}

Where the drift velocity, V_{d} = \frac{L}{t_{d} }

V_{d} = \frac{1.8}{0.608 * 10^{-3} } \\V_{d} = 2960.53 cm/s

and the Electric field strength, E = \frac{V}{l_{s} }

E = 5/2

E = 2.5 V/cm

Mobility of the minority carriers:

\mu_{n} = 2960.53/2.5\\\mu_{n} =1184.21 cm^{2} /V-sec

The electron diffusion coefficient, D_{n} = \frac{(\triangle x)^{2} }{16 t_{d} }

\triangle x = (\triangle t )V_{d}, where Δt = separation of pulse seen in an oscilloscope in time( it should be in micro second range)

\triangle x = \frac{(\triangle t) L}{t_{d} } \\\triangle x = \frac{180*10^{-6} * 1.8}{0.608*10^{-3}  }\\\triangle x =0.533 cm

D_{n} = \frac{0.533^{2} }{16 * 0.608 * 10^{-3} }\\D_{n} = 29.20 cm^2 /s

For the Einstein equation to be satisfied, \frac{D_{n} }{\mu_{n} } = \frac{KT}{q} = 0.025 V

\frac{D_{n} }{\mu_{n} } = \frac{29.20}{1184.21} \\\frac{D_{n} }{\mu_{n} } = 0.025 = 25 mV

Verified.

4 0
2 years ago
2. Harry is pushing a car down a level road at 2.0 m/s with a force of 243 N. The total force
Arte-miy333 [17]

b) Equal to 243 N.

Explanation:

The total force acting on the car in the opposite direction including the road friction and air resistance is equal to 243 N.

This is in conformity with newton's third law of motion.

Newton's third law of motion states that "action and reaction are equal and opposite in direction. "

  • The action force is that of the pull by Harry acting on the car.
  • The reaction force is in the opposite direction.
  • Both action and reaction force equal and opposite and magnitude and direction

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

3 0
2 years ago
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